Please help me solve this radical.

Question

Please show me how does $\sqrt{12\sqrt[3]{2} - 15}$ + $\sqrt{12\sqrt[3]{4} - 12}$ = 3. I can't find the answer anywhere else.

The answer in the book also says $$\left(\sqrt{12\sqrt[3]{2} - 15} + \sqrt{12\sqrt[3]{4} - 12}\right)^2 = 12\sqrt[3]{2} + 12\sqrt[3]{4} - 27 + 2 = 9$$ which I don't see in my calculation. I can understand the $12\sqrt[3]{2} + 12\sqrt[3]{4} - 27$ but where does the $+ 2$ come from? Cuz instead of $+ 2$, I've got $+ 2\sqrt{204-144\sqrt[3]{2}+180\sqrt[3]{4}}$. I don't see how it's the same number.

Answer 1

First, let's make the numbers smaller and the expressions a bit more manageable by dividing through by $\surd12=2\surd3$ and writing $\alpha^3=2$. Then we need to prove that $$\sqrt{\alpha-\tfrac54}+\sqrt{\alpha^2-1}=\tfrac12\surd3.$$ We can do this by demonstrating the equivalent squared equality, which a little simplification reduces to $$\alpha^2+\alpha-\tfrac94+\sqrt{13-4\alpha-5\alpha^2}=(\tfrac12\surd3)^2,$$ or $3-\alpha-\alpha^2=\sqrt{13-4\alpha-5\alpha^2}.$ This identity is easily verified, again, by squaring and simplifying.

Answer 2

Squaring works of course, though there seems to be an error in the book and in your workings as posted above. For ease let us use $x^3=2$. Then we need to show

$$\sqrt{12x-15} + \sqrt{12x^2-12} = 3$$ As both sides are positive, equivalently we show the squares on both sides are the same, viz. $$12x^2+12x-27 + 2\sqrt{(12x-15)(12x^2-12)} = 9$$ $$\iff x^2+x+\sqrt{13-4x-5x^2} = 3$$ We can isolate the radical and once again square to get the equivalent $$13-4x-5x^2 = (3-x-x^2)^2 \iff (x+2)(x^3-2)= 0$$ which is true obviously.

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P.S. Another approach would be to denest the outer radical, if possible.

For e.g. let $\sqrt{12x-15} = a+bx+ cx^2$, then we have $12x-15 = (a^2+4bc)+2(ab+c^2)x + (2ca+b^2)x^2$. Hence we need $2ca=-b^2, ab+c^2=6, a^2+4bc=-15 \implies (a, b, c) = (1, 2, -2)$ for a positive solution. Thus $\sqrt{12\sqrt[3]2-15} = 1 + 2\sqrt[3]2 -2\sqrt[3]4$.

Similarly, $\sqrt{12\sqrt[3]4-12} = 2 - 2\sqrt[3]2+2\sqrt[3]4 $, so the sum is obviously $3$.